Question

a) Discuss the main role of per-unit systems. (CO3:P01 - 3 marks) b) A single phase two winding transformer rated at 20 kVA, 415/230 volts 50 Hz is subjected to open and short circuit test. Table Q3 (a) summarize the results of the test; Table Q3 (a) Open Circuit Test 12 = 12 A P2 = 200 W Short Circuit Test V1 = 35 V P1 = 300 W I. Based on the short circuit test, determine the equivalent series impedance in polar form referred to winding 1 and sketch in s-plane. Neglect the shunt admittance. (C01:PO2 - 4.5 marks) ii. Based on the short circuit test, determine the shunt admittance in polar form referred to winding 1 and sketch in s-plane. Neglect the series impedance. (C01:PO2 - 4.5 marks)

Answer 1

2010 } go) per-cnit system (ond difficult to Whenever we have to deal with fault analges. any other electrical problem analges, we have to deal with larger number of components with their larger values. But į is very with those values so, this perinit System makes as to simplify the problem with Small Sma vahes. Then, Can analse any big networks within Lea tim analge we can and effectively.

bol. S. 20kvA Vlva gated voltase, E415 230 f=5002 O.c set To=IA Por 200W =VIo costo 0.( test shald be conducted at So, vo= 230V Test conducted on secondary side. shoot ciocic test It should be conducted at rated ISC = 20x103 Coorent. Cuis Ise = 83.47A (Vss),- 35V, (Ps], 53000 i) (Zse), (vs.) . (Rs.),= Psc 35 83.470 Isc Asc) 2 (Z) = 0.422 -300 83-47)2 (Rse) 0.063. :. (Ese),? - (Rse)? vo (042²_ @.043)² X se, - 0.4177 N $(25e) 0.043+ 10.4172) = 0.42L84.122

1) (Ro). Vo) Cocow). Coso 0.434 21 230 (0)2 (2 3,0 2x0434 (Ro)2 = 2650 (Ro) - Roz) K2 POD 2 415 Ro. - - 675.140. q 002 230 2x Sin Coos Co.4340) 127.2w -> XL 129-2X41s 2 414.121N , 230 S (2012 675.14+ 3414.121 Gall 675.14-1414.124 695.14+414.721 fol, i 792.03 0.8524-10.523/s Golia 11000051-31.5301s

(son). 5=25km يا م v 1200 V2 7120 124 =0:1 oll Top f = 6oba, ay Ricorn b) R25 0.008N L1:5.305mHD X2=2xTy Gox 5.305 1000 cY YL, = 1.99. d) (X22) = 2 xnx60x 0.05305 1000 XL2= 0.01990 e) toico.003- 70.02 i) Roc Ritra- Rit R2 ka 0.8 +0.008 (0.2 1.6. Rol Xol. X AX21 a 1.99 + 0.0199 02 Xoi - 3.98 ~ 48.9 (0-44834–3 0.9889) 7.335 Zsh <2444.98,-81.47°

Equivalent Golut defied to primar 1 Mem 0.85 til 99 aum RU 0.8 +36th Incen VI = 12000 348.4 V2 7335 79).V = ? V2 = 12ov Cosp=0.8 lag s = 15kVA device, power Constant Transforma a Solvit V2 I2 we know one relation, Vi V2 12 120 » 15710² - 120x72 = I2 = 1500 12-125A | -36.869 125 120 I